Problem: Solve for $x$ : $4x^2 + 8x + 4 = 0$
Solution: Dividing both sides by $4$ gives: $ x^2 + {2}x + {1} = 0 $ The coefficient on the $x$ term is $2$ and the constant term is $1$ , so we need to find two numbers that add up to $2$ and multiply to $1$ The number $1$ used twice satisfies both conditions: $ {1} + {1} = {2} $ $ {1} \times {1} = {1} $ So $(x + {1})^2 = 0$ $x + 1 = 0$ Thus, $x = -1$ is the solution.